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4'x = 8x -----eq1

2'2x = (2³)x

x = 2'2x/2³

x = 2'(2x-3)

sub x in eq1

4'(2'(2x-3)) = 8(2'(2x-3))

16'(2x-3) = 2³(2'(2x-3))

2'(4(2x-3)) = 2'(3+2x-3)

base cancels out

4(2x-3) = 2x

8x - 12 = 2x

6x =12

x = 2 = Final answer!

13-May-2015 22:48:44 ·

2'2x = (2³)x

x = 2'2x/2³

x = 2'(2x-3)

sub x in eq1

4'(2'(2x-3)) = 8(2'(2x-3))

16'(2x-3) = 2³(2'(2x-3))

2'(4(2x-3)) = 2'(3+2x-3)

base cancels out

4(2x-3) = 2x

8x - 12 = 2x

6x =12

x = 2 = Final answer!

13-May-2015 22:48:44 ·

Most Recent Answers (16) |

4^x=8x

soln: 4^x=8x

4^x=x(2^2+2^2)

4^x=x(4)^2

4^x=x(2)^2×2

2^2x=2x^4

2x=4

dividing both sides by 2.

x= 2

soln: 4^x=8x

4^x=x(2^2+2^2)

4^x=x(4)^2

4^x=x(2)^2×2

2^2x=2x^4

2x=4

dividing both sides by 2.

x= 2

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12-May-2015 00:04:26 ·
4^x=8x,

(1 3)^x=8x,

1 3x x(x-1)/2=8x,

1 3x (x^2-x)/2=8x,

(2 6x x^2-x)/2=8x,

2 6x x^2-x=16x,

x^2 6x-x-16x 2=0,

x^2-11x 2=0. Solve the equation quadratically,u will get 2.

(1 3)^x=8x,

1 3x x(x-1)/2=8x,

1 3x (x^2-x)/2=8x,

(2 6x x^2-x)/2=8x,

2 6x x^2-x=16x,

x^2 6x-x-16x 2=0,

x^2-11x 2=0. Solve the equation quadratically,u will get 2.

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12-May-2015 18:26:44 ·
which later gives 9x^2-19x+2=0 as the quadratic equation .

You can't just do maths anyhow. You can verify by put x=3

4'x = 8x -----eq1

2'2x = (2³)x

x = 2'2x/2³

x = 2'(2x-3)

sub x in eq1

4'(2'(2x-3)) = 8(2'(2x-3))

16'(2x-3) = 2³(2'(2x-3))

2'(4(2x-3)) = 2'(3+2x-3)

base cancels out

4(2x-3) = 2x

8x - 12 = 2x

6x =12

x = 2 = Final answer!

2'2x = (2³)x

x = 2'2x/2³

x = 2'(2x-3)

sub x in eq1

4'(2'(2x-3)) = 8(2'(2x-3))

16'(2x-3) = 2³(2'(2x-3))

2'(4(2x-3)) = 2'(3+2x-3)

base cancels out

4(2x-3) = 2x

8x - 12 = 2x

6x =12

x = 2 = Final answer!

If this is the right answer, click "Like" below

13-May-2015 22:48:44 ·
4^x=8x

log2^2x=log8 + logx ( tak log to base 2 )

2xlog2 = 3log2 + logx

2x=3 + logx

2x = 3 + logx/log2 ( change log to base x)

2x = 3 +1/log2 ( log nw in base x)

2x log2/log2= 3 log2/log2 + 1/log2 (as log2/log2=1)

2x log2/log2=(3log2 + 1)/log2 (simplify rhs)

2x log2= 3logx +1

2xlog2 -3log2=1

log2(2x-3)=1 ( rem log is in base x)

2(2x-3)=x

4x-6=×

3x=6

x=2 (as required)

if any mistake , error or mathz error feel free to corect me. nobody is an highland of knwledge

log2^2x=log8 + logx ( tak log to base 2 )

2xlog2 = 3log2 + logx

2x=3 + logx

2x = 3 + logx/log2 ( change log to base x)

2x = 3 +1/log2 ( log nw in base x)

2x log2/log2= 3 log2/log2 + 1/log2 (as log2/log2=1)

2x log2/log2=(3log2 + 1)/log2 (simplify rhs)

2x log2= 3logx +1

2xlog2 -3log2=1

log2(2x-3)=1 ( rem log is in base x)

2(2x-3)=x

4x-6=×

3x=6

x=2 (as required)

if any mistake , error or mathz error feel free to corect me. nobody is an highland of knwledge

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19-May-2015 01:51:00 ·
X=2 if not still contented with cross check by substitute x=2 in the equation.

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19-May-2015 21:12:55 ·
using logbase2.... logbase2 (4^x) =logbase2 (8x) 2xlogbase2 (2)=logbase2 (8)+logbase2 (x) 2x=3logbase2 (2)+logbase2 (x) 2x=3 + logbase2 (x) 2x-3=logbase2 (x) 2^(2x-3)=x^1 2=x and 2x-3=1 [2x-3=1;2x=4;x=2]

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20-May-2015 10:57:43 ·
Please I also need the answer to the above question. Anybody with the correct answer should please help with a solution. Thanks.

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07-Nov-2015 [22:05:54] ·
4^x=8x,

(1+3)^x=8x,

1+3x+x(x-1)3^2/2=8x,

1+3x+(9x^2-9x)/2=8x,

(2+6x+9x^2-9x)/2=8x,

9x^2-3x+2=16x,

9x^2-3x-16x+2=0,

9x^2-19x+2=0,

solve quadratically,

(9x-1)(x-2)=0,

9x-1=0 or x-2=0,

9x=1 or x=2,

x=1/9 or x =2,

therefore x=2(it satisfy the equ.).

Hey guys, to solve that you'll have know a particular rule thats,

LogaN=logbN/logbA.

So, 4'x=8x

XLog(base 2)4=3log(base 2)2 + log(base 2)x. (you'll get that after simplifying)

then,

2x=3 +log(base10)x/log(base10)2

2x=3+x/2

4x=6+x

3x=6, x=2.

LogaN=logbN/logbA.

So, 4'x=8x

XLog(base 2)4=3log(base 2)2 + log(base 2)x. (you'll get that after simplifying)

then,

2x=3 +log(base10)x/log(base10)2

2x=3+x/2

4x=6+x

3x=6, x=2.

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22-May-2016 20:39:35 ·
A 19 year old boy named Ayanlaja Ramoni Adebola from Nigeria has just discovered is own mathematical formula for solving mathematical problem of a^x=bx. A question was asked by his friend to him which brought his research to find the solution to the mathematical problem. He was asked to find the value of x in the equation 4^x=8x.He had to make research on such questions hoping they exist. But he discovered that most people know that the answer was 2 but what is the solution(workings) to the problem? Some made use of graph method, some made use of Eisten Raph method but he noticed most people had to limit the value of 2 because they already knew the answer was 2.

He made his research on such question and concluded there must be workings leading to the solution of the problem.

By his handwork he proposed his own formula which he named "Debo's formula". Which states that for the equation a^x=bx, then x={log{b/loga)} /loga. (where x is approximated to a whole number).

Chek out the solution of 4^x=8x using Debo's formula.

4^x=8x

Solution

a=4 and b=8

Debo's formula ==>x=log{b/loga} divided by log a

X=log{8/log4} /log4

X=log{8/0.6021} /log4

X=log 13.2868/log4

X=1.1234/0.6021

X=1.86=2(Approximate to the nearest whole number)

Thus, using Debo's formula x=2.

Proof that x=2 from Debo's formula

4^x=8x

4^2=8*2

16=16

His friends argued that his formula can only be used to solve 4^x=8x...Another question of that type was created.

Solve 3^x=9x

Using Debo's formula

a=3 and b=9

x=log{b/loga} divided by loga

x=log{9/log3) divided by log3

x=log(9/0.4771) /log3

x=log18. 864/log3

x=1.2756/0.4771

x=2.67=3(approximate to nearest whole number)

X=3 using Debo's formula.

Proof ==>

3^x=9x

3^3=9*3

27=27

Using Debo's formula

Another problem was created

8^x=32x

Using Debo's formula

a=8 and b=32

Debo's formula ==>x=log{b/loga} divided by loga

x=log{32/log8} /log8

x=log {32/0.9031}/log 8

X=log 35.43/log8

X=1.5494/0.903

X=1.7=2(Approximate to the nearest whole number)

Using Debo's formula x=2

Proof that x=2

8^x=32x

8^2=32*3

64=64

The conditions in using Debo's formula are both the LHS and RHS numeric value must be both even or both odd. Also the numeric value at exponential part (a^x) must be lesser and be divisible by the numeric value of the side with identity x. Thus in a^x=bx, a

He made his research on such question and concluded there must be workings leading to the solution of the problem.

By his handwork he proposed his own formula which he named "Debo's formula". Which states that for the equation a^x=bx, then x={log{b/loga)} /loga. (where x is approximated to a whole number).

Chek out the solution of 4^x=8x using Debo's formula.

4^x=8x

Solution

a=4 and b=8

Debo's formula ==>x=log{b/loga} divided by log a

X=log{8/log4} /log4

X=log{8/0.6021} /log4

X=log 13.2868/log4

X=1.1234/0.6021

X=1.86=2(Approximate to the nearest whole number)

Thus, using Debo's formula x=2.

Proof that x=2 from Debo's formula

4^x=8x

4^2=8*2

16=16

His friends argued that his formula can only be used to solve 4^x=8x...Another question of that type was created.

Solve 3^x=9x

Using Debo's formula

a=3 and b=9

x=log{b/loga} divided by loga

x=log{9/log3) divided by log3

x=log(9/0.4771) /log3

x=log18. 864/log3

x=1.2756/0.4771

x=2.67=3(approximate to nearest whole number)

X=3 using Debo's formula.

Proof ==>

3^x=9x

3^3=9*3

27=27

Using Debo's formula

Another problem was created

8^x=32x

Using Debo's formula

a=8 and b=32

Debo's formula ==>x=log{b/loga} divided by loga

x=log{32/log8} /log8

x=log {32/0.9031}/log 8

X=log 35.43/log8

X=1.5494/0.903

X=1.7=2(Approximate to the nearest whole number)

Using Debo's formula x=2

Proof that x=2

8^x=32x

8^2=32*3

64=64

The conditions in using Debo's formula are both the LHS and RHS numeric value must be both even or both odd. Also the numeric value at exponential part (a^x) must be lesser and be divisible by the numeric value of the side with identity x. Thus in a^x=bx, a

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09-Jul-2016 21:13:03 ·
4^x = 8x

Introduce natural logathrims to both sides

Log4^x = log8x

From loga + logb = logab

.: log4^x = log8 + logx

Express the numbers in their simplest form

Log2^2x = log2^3 + logx

apply the rule of power law

2xlog2 = 3log2 + logx

By Rearranging the terms

logx = 2xlog2 - 3log2

by factorizing the expression

Logx = log2(2x - 3)

by dividing through by log2

Logx/log2 = (2x - 3)

Note loga/logb = a/b

Therefore we have,

x/2 = (2x - 3)

Multiply through by 2

x = 4x - 6

By collecting their like term,

-3x = -6

Divide both sides by -3

X = 2 ..........................proved

Introduce natural logathrims to both sides

Log4^x = log8x

From loga + logb = logab

.: log4^x = log8 + logx

Express the numbers in their simplest form

Log2^2x = log2^3 + logx

apply the rule of power law

2xlog2 = 3log2 + logx

By Rearranging the terms

logx = 2xlog2 - 3log2

by factorizing the expression

Logx = log2(2x - 3)

by dividing through by log2

Logx/log2 = (2x - 3)

Note loga/logb = a/b

Therefore we have,

x/2 = (2x - 3)

Multiply through by 2

x = 4x - 6

By collecting their like term,

-3x = -6

Divide both sides by -3

X = 2 ..........................proved

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11-Jul-2016 17:20:15 ·
Thanks Guys bt i have not seen any meaningful proof yet

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07-Nov-2016 22:44:48 ·
4^x=8x

Divide through by x

Then, 4^x/x =8

4^x/x =16/2 since 16/2 =8

Equate either the numerator or denominator

4^x =16 =4^2

X=2

Proved.......

Divide through by x

Then, 4^x/x =8

4^x/x =16/2 since 16/2 =8

Equate either the numerator or denominator

4^x =16 =4^2

X=2

Proved.......

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09-Nov-2016 21:50:17 ·
This question seems to have no true solution...4^x=8x. And if there is I would like to know it.

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18-Nov-2016 15:59:30 ·
Take log of both sides,xlog4=log8x,xlog(2 raise to power 2)=log8+logx,2xlog2=3log2+logx, 2xlog2-3log2=logx,log2(2x-3) = logx,divide both sides by log,2(2x-3)=x,4x-6 = x, 4x-x=6,3x=6,x=(6/3),x=2.

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17-Dec-2016 08:29:00 ·
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